$\displaystyle \int{x^2tan^{-1}xdx} $

$\displaystyle \int{x^2tan^{-1}xdx} = \frac{x^3}{3}tan^{-1}x-\frac{1}{3}\int \frac {x^3}{1+x^2}dx$

$\displaystyle let {}u=1+x^2, \frac{du}{2}=xdx$

$\displaystyle \frac{x^3}{3}tan^{-1}x- \frac{1}{6}\int (1-1/u)$

$\displaystyle \frac{x^3}{3}tan^{-1}x-(x^2/6)-(1/6)+(1/6)ln(x^2+1)$

In the answer, there is no -1/6. Help?